:: In noncommutative mathematics, the Connes embedding problem or conjecture asks whether every finite von Neumann algebra (separable II1 factor) can be embedded into the ultrapower of the hyperfinite II1 factor (countably generated finite factors)[...]
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I rewrote the sentence to read as above. Formerly it saysaid "any finite von Neumann algebra". In this context, "any" is ambiguous. Is it the case that "any finite von Neumann algebra" can be so embedded? If so, then "every" is true. But "asks whether any finite von Neumann algebra" can be so embedded could be construed as: "asks whether there is any finite von Neumann algebra" that can be so embedded. If there is one, then the answer is "yes", even if "every" is not true. [[User:Michael Hardy|Michael Hardy]] ([[User talk:Michael Hardy|talk]]) 01:15, 30 November 2009 (UTC)
